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1 21-1 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 Chapter 21 Electrochemistry: Chemical Change and Electrical Work 21-2

3 Electrochemistry: Chemical Change and Electrical Work 21.1 Half-Reactions and Electrochemical Cells 21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 21.3 Cell Potential: Output of a Voltaic Cell 21.4 Free Energy and Electrical Work 21.5 Electrochemical Processes in Batteries 21.6 Corrosion: A Case of Environmental Electrochemistry 21.7 Electrolytic Cells: Using Electrical Energy to Drive a Nonspontaneous Reaction 21-3

4 Key Points About Redox Reactions Oxidation (electron loss) always accompanies reduction (electron gain). The oxidizing agent is reduced, and the reducing agent is oxidized. The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent. 21-4

5 Figure 21.1 A summary of redox terminology. OXIDATION One reactant loses electrons. Reducing agent is oxidized. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Zn loses electrons. Zn is the reducing agent and becomes oxidized. Oxidation number increases. REDUCTION Other reactant gains electrons. Oxidizing agent is reduced. Oxidation number decreases The oxidation number of Zn increases from x to +2. Hydrogen ion gains electrons. Hydrogen ion is the oxidizing agent and becomes reduced. The oxidation number of H decreases from +1 to 0.

6 Half-Reaction Method for Balancing Redox Reactions Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. Each reaction is balanced for mass (atoms) and charge. One or both are multiplied by some integer to make the number of electrons gained and lost equal. The half-reactions are then recombined to give the balanced redox equation. Advantages: The separation of half-reactions reflects actual physical separations in electrochemical cells. The half-reactions are easier to balance especially if they involve acid or base. It is usually not necessary to assign oxidation numbers to those species not undergoing change. 21-6

7 Balancing Redox Reactions in Acidic Solution Cr 2 O 7 (aq) + I - (aq) Cr 3+ (aq) + I 2 (aq) 1. Divide the reaction into half-reactions - Determine the O.N.s for the species undergoing redox Cr 2 O 7 (aq) + I - (aq) Cr 3+ (aq) + I 2 (aq) Cr 2 O 7 Cr 3+ Cr is going from +6 to +3 I - I 2 I is going from -1 to 0 2. Balance atoms and charges in each half-reaction - 6e H + (aq) + Cr 2 O 7 2 Cr H 2 O(l) 14H + (aq) + net: +12 net: +6 Add 6e - to left. Cr 2 O 7 2 Cr H 2 O(l) 21-7

8 Balancing Redox Reactions in Acidic Solution continued 6e H + (aq) + Cr 2 O 7 2 Cr H 2 O(l) 2 I - I + 2e - 2 Cr(+6) is the oxidizing agent and I(-1) is the reducing agent. 3. Multiply each half-reaction by an integer, if necessary - 2 I - I 2 + 2e - X 3 4. Add the half-reactions together - 6e H + + Cr 2 O 7 2 Cr H 2 O(l) 6 I - 3 I 2 + 6e - 14H + (aq) + Cr 2 O 7 (aq) + 6 I - (aq) 2Cr 3+ (aq) + 3I 2 (s) + 7H 2 O(l) Do a final check on atoms and charges. 21-8

9 Balancing Redox Reactions in Basic Solution Balance the reaction in acid and then add OH - so as to neutralize the H + ions. 14H + (aq) + Cr 2 O 7 (aq) + 6 I - (aq) 2Cr 3+ (aq) + 3I 2 (s) + 14OH - (aq) + 14OH - (aq) + 7H 2 O(l) 14H 2 O + Cr 2 O I - 2Cr I 2 + 7H 2 O + 14OH - Reconcile the number of water molecules. 7H 2 O + Cr 2 O I - 2Cr I OH - Do a final check on atoms and charges. 21-9

10 Figure 21.2 The redox reaction between dichromate ion and iodide ion. Cr 2 O 7 I - Cr 3+ + I

11 Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method PROBLEM: Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO 4 and Na 2 C 2 O 4 in basic solution: MnO 4- (aq) + C 2 O 4 (aq) MnO 2 (s) + CO 3 (aq) PLAN: Proceed in acidic solution and then neutralize with base. SOLUTION: MnO - 4 MnO 2 C 2 O 4 CO MnO - 4 MnO 2 C 2 O 4 2 CO 3 4H + + MnO - 4 MnO 2 + 2H 2 O C 2 O 4 + 2H 2 O 2CO 3 + 4H + +3e - +2e

12 Sample Problem 21.1: continued: Balancing Redox Reactions by the Half-Reaction Method 4H + + MnO 4- +3e - MnO 2 + 2H 2 O C 2 O 4 + 2H 2 O 2CO 3 + 4H + + 2e - 4H + + MnO 4- +3e - X 2 MnO 2 + 2H 2 O C 2 O 4 + 2H 2 O 2CO 3 + 4H + + 2e - X 3 8H + + 2MnO 4- +6e - 2MnO 2 + 4H 2 O 3C 2 O 4 + 6H 2 O 6CO H + + 6e - 8H + + 2MnO 4- +6e - 2MnO 2 + 4H 2 O 3C 2 O 4 + 6H 2 O 6CO H + + 6e - 2MnO (aq) + 3C 2 O 4 (aq) + 2H 2 O(l) 2MnO 2 (s) + 6CO 3 (aq) + 4H + (aq) + 4OH - + 4OH - 2MnO (aq) + 3C 2 O 4 (aq) + 4OH - (aq) 2MnO 2 (s) + 6CO 3 (aq) + 2H 2 O(l) 21-12

13 Figure 21.3 General characteristics of voltaic and electrolytic cells. VOLTAIC CELL System Energy does is released work on from its spontaneous surroundings redox reaction ELECTROLYTIC CELL Surroundings(power Energy is absorbed to supply) drive a nonspontaneous do work system(cell) redox reaction Oxidation half-reaction X X + + e - Reduction half-reaction Y + + e - Y Overall (cell) reaction X + Y + X + + Y; G < 0 Oxidation half-reaction A - A + e - Reduction half-reaction B + + e - B Overall (cell) reaction A - + B + A + B; G > 0

14 Figure 21.4 The spontaneous reaction between zinc and copper(ii) ion Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)

15 Figure 21.5 A voltaic cell based on the zinc-copper reaction. Oxidation half-reaction Zn(s) Zn 2+ (aq) + 2e - Reduction half-reaction Cu 2+ (aq) + 2e - Cu(s) Overall (cell) reaction Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)

16 Notation for a Voltaic Cell components of anode compartment (oxidation half-cell) components of cathode compartment (reduction half-cell) phase of lower oxidation state phase of higher oxidation state phase of higher oxidation state phase of lower oxidation state phase boundary between half-cells Examples: Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) Zn(s) Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e - Cu(s) graphite I - (aq) I 2 (s) H + (aq), MnO 4- (aq) Mn 2+ (aq) graphite inert electrode 21-16

17 Figure 21.6 A voltaic cell using inactive electrodes. Oxidation half-reaction 2I - (aq) I 2 (s) + 2e - Reduction half-reaction MnO 4- (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) Overall (cell) reaction 2MnO 4- (aq) + 16H + (aq) + 10I - (aq) 2Mn 2+ (aq) + 5I 2 (s) + 8H 2 O(l)

18 Sample Problem 21.2: Diagramming Voltaic Cells PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO 3 ) 3 solution, another half-cell with an Ag bar in an AgNO 3 solution, and a KNO 3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. PLAN: SOLUTION: Identify the oxidation and reduction reactions and write each halfreaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction). Oxidation half-reaction Cr(s) Cr 3+ (aq) + 3e - e - Cr Voltmeter salt bridge K + Ag Reduction half-reaction Ag + (aq) + e - Ag(s) Overall (cell) reaction Cr(s) + Ag + (aq) Cr 3+ (aq) + Ag(s) Cr 3+ NO 3 - Ag + Cr(s) Cr 3+ (aq) Ag + (aq) Ag(s)

19 Why Does a Voltaic Cell Work? The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit. E cell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C) 21-19

20 Table 21.1 Voltages of Some Voltaic Cells Voltaic Cell Voltage (V) Common alkaline battery Lead-acid car battery (6 cells = 12V) Calculator battery (mercury) Electric eel (~5000 cells in 6-ft eel = 750V) Nerve of giant squid (across cell membrane)

21 Figure 21.7 Determining an unknown E 0 half-cell with the standard reference (hydrogen) electrode. Oxidation half-reaction Zn(s) Zn 2+ (aq) + 2e - Overall (cell) reaction Zn(s) + 2H 3 O + (aq) Zn 2+ (aq) + H 2 (g) + 2H 2 O(l) Reduction half-reaction 2H 3 O + (aq) + 2e - H 2 (g) + 2H 2 O(l) 21-21

22 Sample Problem 21.3: Calculating an Unknown E 0 half-cell from E0 cell PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br 2 (aq) + Zn(s) Zn 2+ (aq) + 2Br - (aq) E 0 cell = 1.83V Calculate E 0 bromine given E0 zinc = -0.76V PLAN: The reaction is spontaneous as written since the E 0 cell is (+). Zinc is being oxidized and is the anode. Therefore the E 0 bromine can be found using E 0 cell = E0 cathode -E0 anode. SOLUTION: anode: Zn(s) Zn 2+ (aq) + 2e - E = E 0 Zn as Zn 2+ (aq) + 2e - Zn(s) is -0.76V E 0 cell = E0 cathode -E0 anode = 1.83 = E 0 bromine - (-0.76) E 0 bromine = = 1.07 V 21-22

23 Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction E 0 (V) F 2 (g) + 2e - 2F - (aq) strength of oxidizing agent Cl 2 (g) + 2e - 2Cl - (aq) MnO 2 (g) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) Ag + (aq) + e - Ag(s) Fe 3+ (g) + e - Fe 2+ (aq) O 2 (g) + 2H 2 O(l) + 4e - 4OH - (aq) Cu 2+ (aq) + 2e - Cu(s) 2H + (aq) + 2e - H 2 (g) N 2 (g) + 5H + (aq) + 4e - N 2 H 5+ (aq) Fe 2+ (aq) + 2e - Fe(s) 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) Na + (aq) + e - Na(s) strength of reducing agent Li + (aq) + e - Li(s)

24 Writing Spontaneous Redox Reactions By convention, electrode potentials are written as reductions. When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E 0 cell. When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Example: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) stronger reducing agent stronger oxidizing agent weaker oxidizing agent weaker reducing agent 21-24

25 Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E 0 cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents: (1) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) (2) N 2 (g) + 5H + (aq) + 4e - N 2 H 5+ (aq) (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) E 0 = 0.96V E 0 = -0.23V E 0 = 1.23V PLAN: Put the equations together in varying combinations so as to produce (+) E 0 cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E 0. Balance the number of electrons gained and lost without changing the E 0. In ranking the strengths, compare the combinations in terms of E 0 cell

26 Sample Problem 21.4: continued (2 of 4) Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength SOLUTION: (1) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) E 0 = 0.96V (a) Rev (2) N 2 H 5+ (aq) N 2 (g) + 5H + (aq) + 4e - E 0 = +0.23V (1) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) X4 (2) N 2 H 5+ (aq) N 2 (g) + 5H + (aq) + 4e - X3 E 0 cell = 1.19V (A) 4NO 3- (aq) + 3N 2 H 5+ (aq) + H + (aq) 4NO(g) + 3N 2 (g) + 8H 2 O(l) Rev (1) NO(g) + 2H 2 O(l) NO 3- (aq) + 4H + (aq) + 3e - (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) (1) NO(g) + 2H 2 O(l) NO 3- (aq) + 4H + (aq) + 3e - X2 (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) X3 E 0 = -0.96V E 0 = 1.23V E 0 cell = 0.27V (B) NO(g) + 3MnO 2 (s) + 4H + (aq) 2NO 3- (aq) + 3Mn 3+ (aq) + 2H 2 O(l)

27 Sample Problem 21.4: continued (3 of 4) Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength Rev (2) N 2 H 5+ (aq) N 2 (g) + 5H + (aq) + 4e - (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) (2) N 2 H 5+ (aq) N 2 (g) + 5H + (aq) + 4e - (3) MnO 2 (s) +4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) X2 E 0 = +0.23V E 0 = 1.23V E 0 cell = 1.46V (C) N 2 H 5+ (aq) + 2MnO 2 (s) + 3H + (aq) N 2 (g) + 2Mn 2+ (aq) + 4H 2 O(l) (b) Ranking oxidizing and reducing agents within each equation: (A): oxidizing agents: NO 3- > N 2 reducing agents: N 2 H 5+ > NO (B): oxidizing agents: MnO 2 >NO 3 - reducing agents: NO > Mn 2+ (C): oxidizing agents: MnO 2 > N 2 reducing agents: N 2 H 5+ > Mn 2+

28 Sample Problem 21.4: continued (4 of 4) Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of Oxidizing agents: MnO 2 > NO 3- > N 2 Reducing agents: N 2 H 5+ > NO > Mn

29 Relative Reactivities (Activities) of Metals 1. Metals that can displace H from acid 2. Metals that cannot displace H from acid 3. Metals that can displace H from water 4. Metals that can displace other metals from solution 21-29

30 Figure 21.8 The reaction of calcium in water. Oxidation half-reaction Ca(s) Ca 2+ (aq) + 2e - Reduction half-reaction 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) Overall (cell) reaction Ca(s) + 2H 2 O(l) Ca 2+ (aq) + H 2 (g) + 2OH - (aq) 21-30

31 Free Energy and Electrical Work G α -E cell -E cell = -w max charge charge = n F n = #mols e - F = Faraday constant F = 96,485 C/mol e - G = w max = charge x (-E cell ) G = -n F E cell In the standard state - G 0 = -n F E 0 cell G 0 = - RT ln K E 0 cell = - (RT/n F) ln K 1V = 1J/C F = 9.65x10 4 J/V*mol e

32 Figure 21.9 The interrelationship of G 0, E 0, and K. G 0 G 0 K E 0 cel l Reaction at standard-state conditions < 0 > 1 > 0 spontaneous at equilibrium > 0 < 1 < 0 nonspontaneous G 0 = -nfe o cell G 0 = -RT lnk E 0 cel K By substituting standard state values into E 0 cell, we get E 0 cell = (0.0592V/n) log K (at 298 K) l E 0 cell = -RT lnk nf

33 Sample Problem 21.5: Calculating K and G 0 from E 0 cell PROBLEM: Lead can displace silver from solution: Pb(s) + 2Ag + (aq) Pb 2+ (aq) + 2Ag(s) As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and G 0 at 298 K for this reaction. PLAN: Break the reaction into half-reactions, find the E 0 for each half-reaction and then the E 0 cell. Substitute into the equations found on slide SOLUTION: Pb 2+ (aq) + 2e - Ag + (aq) + e - Pb(s) Ag(s) E 0 = -0.13V E 0 = 0.80V E 0 = 0.13V E 0 = 0.80V 2X Pb(s) Pb 2+ (aq) + 2e - Ag + (aq) + e - Ag(s) E 0 cell = 0.93V E 0 cell = 0.592V n log K G 0 = -nfe 0 cell = -(2)(96.5kJ/mol*V)(0.93V) n x E log K = 0 cell (2)(0.93V) = K = 2.6x10 31 G 0 = -1.8x10 2 kj 0.592V 0.592V 21-33

34 The Effect of Concentration on Cell Potential G = G 0 + RT ln Q -nf E cell = -nf E cell + RT ln Q E cell = E 0 cell - RT nf ln Q When Q < 1 and thus [reactant] > [product], lnq < 0, so E cell > E 0 cell When Q = 1 and thus [reactant] = [product], lnq = 0, so E cell = E 0 cell When Q >1 and thus [reactant] < [product], lnq > 0, so E cell < E 0 cell E cell = E 0 cell n log Q 21-34

35 Sample Problem 21.6: Using the Nernst Equation to Calculate E cell PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn 2+ half-cell and an H 2 /H + half-cell under the following conditions: [Zn 2+ ] = 0.010M [H + ] = 2.5M P = 0.30atm H 2 Calculate E cell at 298 K. PLAN: Find E 0 cell and Q in order to use the Nernst equation. SOLUTION: Determining E 0 cell : 2H + (aq) + 2e - H 2 (g) E 0 = 0.00V Zn 2+ (aq) + 2e - Zn(s) E 0 = -0.76V Zn(s) Zn 2+ (aq) + 2e - E 0 = +0.76V E cell = E 0 cell V n log Q E cell = (0.0592/2)log(4.8x10-4 ) = 0.86V Q = Q = P x [Zn 2+ ] H 2 [H + ] 2 (0.30)(0.010) (2.5) 2 Q = 4.8x

36 Figure Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The relation between E cell and log Q for the zinc-copper cell

37 Figure A concentration cell based on the Cu/Cu 2+ half-reaction. Oxidation half-reaction Cu(s) Cu 2+ (aq, 0.1M) + 2e - Reduction half-reaction Cu 2+ (aq, 1.0M) + 2e - Cu(s) Overall (cell) reaction Cu 2+ (aq,1.0m) Cu 2+ (aq, 0.1M)

38 Sample Problem 21.7: Calculating the Potential of a Concentration Cell PROBLEM: A concentration cell consists of two Ag/Ag + half-cells. In half-cell A, electrode A dips into M AgNO 3 ; in half-cell B, electrode B dips into 4.0x10-4 M AgNO 3. What is the cell potential at 298 K? Which electrode has a positive charge? PLAN: E 0 cell will be zero since the half-cell potentials are equal. E cell is calculated from the Nernst equation with half-cell A (higher [Ag + ]) having Ag + being reduced and plating out, and in half-cell B Ag(s) will be oxidized to Ag +. SOLUTION: Ag + (aq, 0.010M) half-cell A Ag + (aq, 4.0x10-4 M) half-cell B E cell = E 0 cell V 1 log [Ag+ ] dilute [Ag + ] concentrated E cell = 0 V log 4.0x10-2 = V Half-cell A is the cathode and has the positive electrode

39 Figure The laboratory measurement of ph. Glass electrode Pt Reference (calomel) electrode AgCl on Ag on Pt 1M HCl Hg Paste of Hg 2 Cl 2 in Hg KCl solution Thin glass membrane Porous ceramic plugs

40 Table 21.3 Some Ions Measured with Ion-Specific Electrodes Species Detected NH 3 /NH + 4 CO 2 /HCO - 3 F - Br - I - NO - 3 K + H + Typical Sample Industrial wastewater, seawater Blood, groundwater Drinking water, urine, soil, industrial stack gases Grain, plant tissue Milk, pharmaceuticals Soil, fertilizer, drinking water Blood serum, soil, wine Laboratory solutions, soil, natural waters 21-40

41 Figure Alkaline Battery 21-41

42 Figure Mercury and Silver (Button) Batteries 21-42

43 Figure Lithium battery

44 Figure Lead-acid battery

45 Figure Nickel-metal hydride (Ni-MH) battery 21-45

46 Figure Lithium-ion battery

47 Figure Fuel cell

48 Figure The corrosion of iron

49 Figure Enhanced corrosion at sea

50 Figure Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The effect of metal-metal contact on the corrosion of iron. faster corrosion cathodic protection 21-50

51 Figure The use of sacrificial anodes to prevent iron corrosion

52 Figure The tin-copper reaction as the basis of a voltaic and an electrolytic cell. voltaic cell electrolytic cell Oxidation half-reaction Sn(s) Sn 2+ (aq) + 2e Reduction half-reaction Cu 2+ (aq) + 2e - Cu(s) Overall (cell) reaction Sn(s) + Cu 2+ (aq) Sn 2+ (aq) + Cu(s) Oxidation half-reaction Cu(s) Cu 2+ (aq) + 2e - Reduction half-reaction Sn 2+ (aq) + 2e - Sn(s) Overall (cell) reaction Sn(s) + Cu 2+ (aq) Sn 2+ (aq) + Cu(s)

53 Figure The processes occurring during the discharge and recharge of a lead-acid battery. VOLTAIC(discharge) ELECTROLYTIC(recharge) 21-53

54 Table 21.4 Comparison of Voltaic and Electrolytic Cells Electrode Cell Type G E cell Name Process Sign Voltaic < 0 > 0 Anode Oxidation - Voltaic < 0 > 0 Cathode Reduction + Electrolytic > 0 < 0 Anode Oxidation + Electrolytic > 0 < 0 Cathode Reduction

55 Sample Problem 21.8: PROBLEM: Predicting the Electrolysis Products of a Molten Salt Mixture A chemical engineer melts a naturally occurring mixture of NaBr and MgCl 2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced halfreactions and the overall cell reaction. PLAN: Consider the metal and nonmetal components of each compound and then determine which will recover electrons(be reduced; strength as an oxidizing agent) better. This is the converse to which of the elements will lose electrons more easily (lower ionization energy). SOLUTION: Possible oxidizing agents: Na +, Mg 2+ Possible reducing agents: Br -, Cl - Na, the element, is to the left of Mg in the periodic table, therefore the IE of Mg is higher than that of Na. So Mg 2+ will more easily gain electrons and is the stronger oxidizing agent. Br, as an element, has a lower IE than does Cl, and therefore will give up electrons as Br - more easily than will Cl -. cathode Mg 2+ (l) + 2Br - (l) anode Mg(s) + Br 2 (g)

56 Figure The electrolysis of water. Overall (cell) reaction 2H 2 O(l) H 2 (g) + O 2 (g) Oxidation half-reaction 2H 2 O(l) 4H + (aq) + O 2 (g) + 4e - Reduction half-reaction 2H 2 O(l) + 4e - 2H 2 (g) + 2OH - (aq) 21-56

57 Sample Problem 21.9: Predicting the Electrolysis Products of Aqueous Ionic Solutions PROBLEM: What products form during electrolysis of aqueous solution of the following salts: (a) KBr; (b) AgNO 3 ; (c) MgSO 4? PLAN: Compare the potentials of the reacting ions with those of water, remembering to consider the 0.4 to 0.6V overvoltage. The reduction half-reaction with the less negative potential, and the oxidation halfreaction with the less positive potential will occur at their respective electrodes. SOLUTION: (a) K + (aq) + e - K(s) E 0 = -2.93V 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) E 0 = -0.42V The overvoltage would make the water reduction to but the reduction of K + is still a higher potential so H 2 (g) is produced at the cathode. 2Br - (aq) Br 2 (g) + 2e - E 0 = 1.07V H 2 O(l) O 2 (g) + 4H + (aq) + 4e - E 0 = 0.82V The overvoltage would give the water half-cell more potential than the Br -, so the Br - will be oxidized. Br 2 (g) forms at the anode.

58 Sample Problem 21.9: continued Predicting the Electrolysis Products of Aqueous Ionic Solutions (b) Ag + (aq) + e - 2H 2 O(l) + 2e - Ag(s) H 2 (g) + 2OH - (aq) E 0 = -0.80V E 0 = -0.42V Ag + is the cation of an inactive metal and therefore will be reduced to Ag at the cathode. Ag + (aq) + e - Ag(s) The N in NO 3- is already in its most oxidized form so water will have to be oxidized to produce O 2 at the anode. 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e - (c) Mg 2+ (aq) + 2e - Mg(s) E 0 = -2.37V Mg is an active metal and its cation cannot be reduced in the presence of water. So as in (a) water is reduced and H 2 (g) is produced at the cathode. The S in SO 4 is in its highest oxidation state; therefore water must be oxidized and O 2 (g) will be produced at the anode

59 Figure Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A summary diagram for the stoichiometry of electrolysis. MASS MASS (g) (g) of of substance oxidized or or reduced M(g/mol) AMOUNT (MOL) (MOL) of of substance oxidized or or reduced AMOUNT (MOL) (MOL) of of electrons transferred Faraday constant (C/mol e - ) balanced half-reaction CHARGE (C) (C) time(s) CURRENT (A) (A) 21-59

60 Sample Problem 21.10: Applying the Relationship Among Current, Time, and Amount of Substance PROBLEM: A technician is plating a faucet with 0.86 g of Cr from an electrolytic bath containing aqueous Cr 2 (SO 4 ) 3. If 12.5 min is allowed for the plating, what current is needed? PLAN: mass of Cr needed SOLUTION: Cr 3+ (aq) + 3e - Cr(s) divide by M mol of Cr needed 3mol e - /mol Cr mol of e - transferred 9.65x10 4 C/mol e - charge (C) divide by time current (A) 0.86g (mol Cr) (3 mol e - ) = mol e - (52.00 gcr) (mol Cr) mol e - (9.65x10 4 C/mol e - ) = 4.8x10 3 C 4.8x10 3 C 12.5 min (min) (60s) = 6.4C/s = 6.4 A 21-60

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